(1) is derived from f (x)+f (y) = x2+2x-3+y2+2y-3 = (x+1) 2+(y+1) 2-8 ≤ 0.

M = {(x, y) | (x+1) 2+(y+1) 2 ≤ 8}, which means a disk with a center of (-1,-1) and a radius of 2√2;

f(x)-f(y)=(x2+2x-3)-(y2+2y-3)=(x-y)(x+y+2)≥0。

N = {(x, y) | (x-y) (x+y+2) ≥ 0}, which represents the left and right two areas in the four areas where straight lines x-y=0 and x+y+2=0 intersect, and the area corresponding to M∩N is the shaded part as shown in the figure.

Therefore, the area = 8π/2 = 4π.

(2)b/(a-3) indicates the slope of the connecting line between point (a, b) and point (3, 0) in the region,

As can be seen from the figure, when the straight line intersects (1, 1), the minimum value is1(1-3) =-1/2.

When the straight line is tangent to the circle, the maximum value is obtained.

Let the linear equation be y = k(x-3) and the distance from the center of the circle to the straight line be | k (-1-3)+1|/√ (k2+1) = 2 √ 2.

The solution is k = (2+3√2)/4 (excluding (2-3√2)/4).

So the range of the value is [- 1/2, (2+3√2)/4].